How do you show that $(sec^2x + csc^2x)(sec^2x- csc^2x) =tan^2x -cot^2x$ ?

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Answer 1 · 2016-12-18T17:00:29

The identity is false.

$(sec^2x + csc^2x)(sec^2x - csc^2x) = tan^2x - cot^2x$

$(1/cos^2x + 1/sin^2x)(1/cos^2x- 1/sin^2x) = sin^2x/cos^2x - cos^2x/sin^2x$

$((sin^2x + cos^2x)/(cos^2xsin^2x))((sin^2x - cos^2x)/(cos^2xsin^2x)) = sin^2x/cos^2x - cos^2x/sin^2x$

$(sin^2x- cos^2x)/(cos^4xsin^4x) = (sin^4x - cos^4x)/(sin^2xcos^2x)$

$(sin^2x -cos^2x)/(cos^4xsin^4x) = ((sin^2x+ cos^2x)(sin^2x - cos^2x)/(sin^2xcos^2x)$

$(sin^2x - cos^2x)/(cos^4xsin^4x) = (sin^2x - cos^2x)/(sin^2xcos^2x)$

As you can see, the two sides are unequal, so the identity is false.

Hopefully this helps!

How do you show that (sec^2x + csc^2x)(sec^2x- csc^2x) =tan^2x -cot^2x(sec^2x + csc^2x)(sec^2x- csc^2x) =tan^2x -cot^2x?